Re: a difficult homework

[shivax]

now i try to expian better
NEW EXAMPLE:

a= Matrix of all equity 

10  4   13   20 
25  -6  10   15
13  10   8   17
8   8   15   22

b and c is a matrix position of a

 c=
    1    5    9   13
    2    6   10   14
    3    7   11   15
    4    8   12   16

 b=
    1    5    9   13
    2    5    9   13
    2    6   9   14
    3    7   11   15


 c: is a matrix of "a" linear position  
    1 a(1,1)   5 a(1,2)    9 a(1,3)   13 a(1,4)
    2 a(2,1)   6 a(2,2)   10 a(2,3)   14 a(2,4)
    3 a(3,1)   7 a(3,2)   11 a(3,3)   15 a(3,4)
    4 a(4,1)   8 a(4,2)   12 a(4,3)   16 a(4,4)

 
 b: is a matrix of "a" linear position
   1 a(1,1)   5 a(1,2)    9 a(1,3)   13 a(1,4)
    2 a(2,1)   5 a(2,2)   9 a(2,3)    13 a(2,4)
    2 a(3,1)   6 a(3,2)   9 a(3,3)   14 a(3,4)
    3 a(4,1)   7 a(4,2)   11 a(4,3)   15 a(4,4)

I thought of creating the matrix c to analyze the minimum of the equity
between the 2 points

why do I need the minimums between the various points of the matrix a?

because a is the equity matrix..and I want to find the minimum peak between
the various surveys (positions) using matrix b and matrix c

- the elements of b are less than a
-the element of b and c are egual/lower of length(a)
 (because are position of a)
- a is a matrix so large..
the matrix a is fed from time therefore the dimension is variable.
However at present it occupies 5000 (row) * 88 (columns) size
I don't think it can exceed 10,000 * 500

that's all ... I think I was exhaustive
(




55/5000


I hope I didn't make mistakes in rebuilding the matrices)





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